Mathematical Tools for Physics

14—Complex Variables 443

Take the radius of the circle large enough that only the first term in the numerator and the first term in the
denominator are important. That makes the integral

∫

C

dz

n

z

= 2πin

It’s certainly not zero, so that means that there is a pole inside the loop, and so a root of the denominator.
Function Determined by its Boundary Values:If a function is analytic throughout a simply connected domain
andCis a simple closed curve in this domain, then the values offinsideCare determined by the values offon
C. Letzbe a point inside the contour then I will show

1

2 πi

∫

C

dz

f(z′)

z′−z

=f(z) (15)

Becausefis analytic in this domain I can shrink the contour to be an arbitrarily small curveC 1 aroundz. Because
fis continuous, I can make the curve close enough toz thatf(z′) =f(z)to any accuracy that I want. That
implies that the above integral is the same as

1

2 πi

f(z)

∫

C 1

dz′

1

z′−z

=f(z)

Eq. ( 15 ) is Cauchy’s integral formula, giving the analytic function in terms of its boundary values.
Derivatives:You can differentiate Cauchy’s formula any number of times.

dnf(z)

dzn

=

n!

2 πi

∫

C

dz

f(z′)

(z′−z)n+1

Entire Functions:An entire function is one that has no singularities anywhere. ez, polynomials, sines, cosines
are such. There’s a curious and sometimes useful result about such functions. A bounded entire function is
necessarily a constant. For a proof, take two points,z 1 andz 2 and apply Cauchy’s integral theorem.

f(z 1 )−f(z 2 ) =

1

2 πi

∫

C

dz f(z′)

[

1

z′−z 1

−

1

z′−z 2

]

=

1

2 πi

∫

C

dz f(z′)

z 1 −z 2

(z′−z 1 )(z′−z 2 )